Question

How do you find the following series, from n=0 to infinity of (1)^n / n!

Answer 1

\(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty } ~ \frac{1^n}{n!}}\) is the same thing as \(\large\color{black}{ \displaystyle \sum_{ n=0 }^{ \infty } ~ \frac{1}{n!}}\)

Answer 2

PLease help me @SolomonZelman

Answer 3

this should converge quite rapidly (we don't even need to show why, but if you want you can use a ratio test)

Answer 4

Here are some partial sums.

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\(\color{black}{ \displaystyle {\rm S}_0=\frac{1}{0!}=1}\) (you know that factorial of 0 is 1)
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\(\color{black}{ \displaystyle {\rm S}_1=\frac{1}{0!}+\frac{1}{1!}=1+1=2}\) (factorial of 1 is also 1)
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\(\color{black}{ \displaystyle {\rm S}_2=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}=1+1+\frac{1}{2}=2+\frac{1}{2} =\color{red}{2.5}}\)
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\(\color{black}{ \displaystyle {\rm S}_3=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}=1+1+\frac{1}{2}+\frac{1}{6}=2+\frac{3}{6}+\frac{1}{6}=2+\frac{4}{6}=2+\frac{2}{3} \approx \color{red}{2.67} }\)
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from here, I will be a little bit more brief.

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\(\color{black}{ \displaystyle {\rm S}_4=\left({\rm S}_3\right)+\left(a_{4}\right)=\left(2+\frac{2}{3}\right)+\left(\frac{1}{4!}\right) =2+\frac{2}{3}+\frac{1}{24}=2+\frac{17}{24}\approx\color{red}{2.71} }\)
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\(\color{black}{ \displaystyle {\rm S}_5=\left({\rm S}_4\right)+\left(a_{5}\right)=\left(2+\frac{17}{24}\right)+\left(\frac{1}{5!}\right) =2+\frac{17}{24}+\frac{1}{120}=2+\frac{43}{60}\approx\color{red}{2.72} }\)
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\(\color{black}{ \displaystyle {\rm S}_6=\left({\rm S}_5\right)+\left(a_{6}\right)=\left(2+\frac{43}{60}\right)+\left(\frac{1}{6!}\right) =2+\frac{43}{60}+\frac{1}{720}=2+\frac{517}{720}\approx\color{red}{2.72} }\)
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so approximately, that is where the series "stops"
after that, the numbers you add are so insignificant!